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What is $$\lim_{x \rightarrow 0} x^0$$? Would this equal to $\lim_{x \rightarrow 0}x^x = 1$?

If the limit is undefined, would $\lim_{x \rightarrow 0^+} x^0$ be defined?

limito
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  • what is this question? $x^0 = 1$ as x approaches 0 $x^0$ is equal to 1 – Irrational Person Dec 25 '14 at 05:27
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    Excellent discussion of different viewpoints here http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/ – Richard Dec 25 '14 at 05:32
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    @Richard I dislike this opening sentence for the mathematicians side: "Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true." There are solid reasons mathematicians choose to define it as $1$. In particular, when dealing with natural numbers, it makes absolute and simple sense to define $x^y$ inductively as $x^0=1$ and $x^{n+1}=x\cdot x^n$. To leave $0^0$ undefined leads to complexity that is pointless. Also, all empty products are $1$, again, because that is the simplest definition. An sequence of no $0$s is no different from a sequence of no 2s. – Thomas Andrews Dec 25 '14 at 05:49
  • @Richard This question is not about $0^0$. But it's natural to remark on the latter anyway, so I'll refer to Zero to the zero power - Is $0^0=1$? which offers an explanation that's both more transparent and more concise. –  Dec 25 '14 at 05:57
  • @ThomasAndrews I agree there are good reasons, but it has not consistently been the mathematicians choice. I recently read a paper on from less than 20 years ago that addressed the argument. The webpage gives some of the reasons for the current near consensus near the end of the article. For me, the binomial theorem was very persuasive evidence that it is the correct choice. – Richard Dec 25 '14 at 06:18
  • This question is relevant; I proved that $\lim_{x\rightarrow 0^+}x^{\alpha x}=1$ in my answer to it. – Milo Brandt Dec 25 '14 at 06:21
  • Lots of people telling me about stuff they've read, nobody making an actual argument. @Richard – Thomas Andrews Dec 25 '14 at 06:28

4 Answers4

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$x^0=1$ for all $x$, in particular for $x\neq 0$. So this is the same as $\lim_{x\to 0} 1 = 1$.

$x^x$ is not well-defined on any $(-\delta,\delta)\setminus\{0\}$, so you have to be careful. $\lim_{x\to 0^+} x^x$ does exist.

Thomas Andrews
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  • But isn't $0^0$ indeterminate? So it's not true that for all $x$, $x^0 =1$... – limito Dec 25 '14 at 05:31
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    But you are not considering $0^0$. The limit explains the behavior around the neighborhood of the point, but not at the point exactly. – MathMajor Dec 25 '14 at 05:33
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    Indeterminate does not mean undefined. It just means that $\lim_{(x,y)\to (0,0)} x^y$ does not exist. We use $x^0=1$ all the time in mathematics - for example, when we write $p(x)=\sum_{i=0}^n a_ix^i$ for a polynomial and say $p(0)=a_0$. – Thomas Andrews Dec 25 '14 at 05:33
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    In particular, in lambda calculus and set theory definitions of $x^y$, we get $0^0=1$ unless we go out of our way to say it is undefined. You get no contradiction by defining $0^0=1$, and we use it quite a bit, so we define it that way. We just have to remember that the function $x^y$ is not continuous there. @limito – Thomas Andrews Dec 25 '14 at 05:34
  • I occasionally define addition so that $1+1 = 0$. That does not mean addition is ambiguous: it means that there are two different notions of addition that I am prone to using. The only ambiguity is which notion of addition I mean when I write $+$. It's the same deal with $0^0$. –  Dec 25 '14 at 05:45
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    @Hurkyl True, but we actually use $0^0=1$ implicitly quite often. It is almost always pointless to make it undefined, as long as we recognize the discontinuity. – Thomas Andrews Dec 25 '14 at 05:52
  • The first sentence would leave less room to doubt (and follow-up questions in comments) if it said "$x^0=1$ for $x\ne 0$". The value of $0^0$ is an issue that occupies a whole other question on this site, and it's not relevant here. –  Dec 25 '14 at 05:52
  • True, but that would perpetuate misinformation (it wouldn't actually be wrong, just imply a common misconception.) @Behaviour – Thomas Andrews Dec 25 '14 at 06:05
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(i) $$\lim_{x \to 0} x^0 = 1$$

because $a^0 = 1$ for any $a \neq 0$. But with the limit, we are considering the neighborhood of $0$, and not $0$ itself.

(ii)

$$\lim_{x \to 0^+} x^x = \exp \left( \lim_{x \to 0^+} x \log x \right) \stackrel{\mathcal{L}}{=} \exp(0)=1 .$$

MathMajor
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$$x^0=1$$was based on the assumption that $x\neq 0$ and only then we could show that $$\frac{x^m}{x^m}=x^{m-m}=x^0=1$$ Therefore since when x goes to 0 x is never 0 then $x^0$goes to 1.

math424
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Answer:

$$\lim_{x \to 0} x^0 = 1 $$

because anything to the zero power will equal 1.

/edit

$\lim_{x\to 0^+} x^x$ does exist.

Irrational Person
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  • You mean anything $\neq 0$? – MathMajor Dec 25 '14 at 05:41
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    But $\lim_{x \to 0^+} x^x$ does exist. It equals one. Use Log-Exp technique to see this from right hand side. – MathMajor Dec 25 '14 at 05:42
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    Yes, @Bot is wrong about the limit of $x^x$. Not clear why he put quotes around "exist," there. – Thomas Andrews Dec 25 '14 at 05:42
  • I'm very skeptical about $0^0$, there was a big thread about this a few days ago with other indeterminate forms and the general consensus was that it was undefined ... ? – MathMajor Dec 25 '14 at 05:43
  • The general consensus is wrong. Mathematicians define it as $1$. Sorry if that bugs you, but that is the definition. Again, see my comments below. – Thomas Andrews Dec 25 '14 at 05:44
  • I'm not sure if I agree with that. Anyhow, I have up voted your answer regardless. I think you did a better job answering it then I did. – MathMajor Dec 25 '14 at 05:48
  • @GabrielH Find me a mathematician who is not talking about estimates for reals numbers (where $0$ really means an interval containing $0$) who leaves $0^0$ undefined. It really shouldn't be hard. I've pointed out several ways that we implicitly define $0^0$ for natural numbers, and why we do so. (An empty list of $0$s is no different from an empty list of $3$s, so $0^0=3^0$.) – Thomas Andrews Dec 25 '14 at 05:55
  • (And no, calling it indeterminate is not the same as calling it undefined.) – Thomas Andrews Dec 25 '14 at 05:56
  • Okay, can we settle on $0^0 \ge \le 1$ where $\ge \le$ denotes maybe? :-) – MathMajor Dec 25 '14 at 05:59
  • Nope, it is equal to $1$. – Thomas Andrews Dec 25 '14 at 05:59
  • I'm just messing with you. Okay, okay, $0^0 = 1$. On the record. – MathMajor Dec 25 '14 at 06:00